/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 The molarity of a solution obtai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 13

The molarity of a solution obtained by mixing \(750 \mathrm{~mL}\) of \(0.5(\mathrm{M}) \mathrm{HCl}\) with \(250 \mathrm{~mL}\) of \(2(\mathrm{M}) \mathrm{HCl}\) will be : (a) \(0.875 \mathrm{M}\) (b) \(1.00 \mathrm{M}\) (c) \(1.75 \mathrm{M}\) (d) \(0.975 \mathrm{M}\)

Short Answer

Expert verified
The molarity of the mixed solution is \(0.875 \mathrm{M}\), option (a).

Step by step solution

01

Calculate moles of HCl from each solution

First, calculate the moles of HCl present in each solution using the formula \( ext{moles} = ext{Molarity} \times ext{Volume (L)} \). For the first solution \( (0.5 \, M, 750 \, mL) \): \( 0.5 \, ext{M} \times 0.750 \, ext{L} = 0.375 \, ext{moles} \). For the second solution \( (2.0 \, M, 250 \, mL) \): \( 2 \, ext{M} \times 0.250 \, ext{L} = 0.5 \, ext{moles} \).
02

Calculate total moles of HCl

Add the moles from both solutions to get the total moles of HCl: \( 0.375 \, ext{moles} + 0.5 \, ext{moles} = 0.875 \, ext{moles} \).
03

Calculate total volume of the mixed solution

Add the volumes of the two solutions to find the total volume: \( 750 \, ext{mL} + 250 \, ext{mL} = 1000 \, ext{mL} \). Convert this to liters: \( 1000 \, ext{mL} = 1.0 \, ext{L} \).
04

Calculate the molarity of the mixed solution

The molarity of the mixed solution is given by \( \text{Molarity} = \frac{\text{Total moles}}{\text{Total volume (L)}} \). Substitute the values: \( \frac{0.875 \, ext{moles}}{1.0 \, ext{L}} = 0.875 \, M \).
05

Match with given options

Compare the calculated molarity \( 0.875 \, M \) with the provided options. The answer is option (a): \( 0.875 \, ext{M} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solution Mixing
When it comes to mixing chemical solutions, it is important to understand how different concentrations of solutions will behave once combined. In this exercise, you are presented with two solutions of hydrochloric acid ( HCl ) that need to be mixed together.
The concentration of a solution is expressed in terms of molarity, which is defined as moles of solute per liter of solution. Each solution has a specific molarity and volume. This exercise involves mixing a 750 mL solution of HCl with a molarity of 0.5 M, and another 250 mL solution with a molarity of 2.0 M. When these two solutions are mixed, the resulting solution will have its own unique molarity that's determined by the combined volumes and moles from each of the original solutions. It is important to note that the chemical identity of the solutions remains unchanged; only the concentration and volume properties are combined.
Chemical Moles Calculation
Calculating moles is central to understanding the composition of a solution. A mole is a unit that quantifies the amount of substance. The key formula for calculating the moles in a solution is: \[ \text{moles} = \text{Molarity} \times \text{Volume (L)} \]
For the first solution in our problem, which has a volume of 750 mL (or 0.750 L) and a molarity of 0.5 M, the moles are calculated as: \[ 0.5 \, \text{M} \times 0.750 \, \text{L} = 0.375 \, \text{moles} \]For the second solution that has a volume of 250 mL (or 0.250 L) and a molarity of 2.0 M, the moles are:\[ 2 \, \text{M} \times 0.250 \, \text{L} = 0.5 \, \text{moles} \]Once the moles of each solution are determined, they are summed to find the total moles present in the final mixed solution. This total will then be used to find the new molarity of the mixture.
Volume Conversion
Volume conversion is a crucial step in calculations involving molarity. Typically, volumes given in milliliters (mL) need to be converted to liters (L) because molarity is expressed in moles per liter.
  • 1,000 mL equals 1.0 L.
  • To convert from mL to L, divide the volume in mL by 1,000.
In our exercise, the volumes of 750 mL and 250 mL need to be added to find the total volume of the mixed solution. Together they make: \[ 750 \, \text{mL} + 250 \, \text{mL} = 1,000 \, \text{mL} \]When converted to liters, this total volume becomes 1.0 L. Accurate volume conversion is important because it significantly impacts the calculation of molarity for the mixed solution. In the end, it's the total moles divided by this converted volume that gives the final molarity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An azeotropic solution of two liquids has boiling point lower than either of them when it (a) shows negative deviation from Raoult's law (b) shows no deviation from Raoult's law (c) shows positive deviation from Raoult's law (d) is saturated

The Henry's law constant for the solubility of \(\mathrm{N}_{2}\) gas in water at \(298 \mathrm{~K}\) is \(1.0 \times 10^{5}\) atm. The mole fraction of \(\mathrm{N}_{2}\) in air is \(0.8\). The number of moles of \(\mathrm{N}_{2}\) from air dissolved in 10 moles of water at \(298 \mathrm{~K}\) and 5 atm pressure is (a) \(4.0 \times 10^{-4}\) (b) \(4.0 \times 10^{5}\) (c) \(5.0 \times 10^{-4}\) (d) \(4.0 \times 10^{-6}\)

On dissolving \(0.5 \mathrm{~g}\) of a non-volatile non-ionic solute to \(39 \mathrm{~g}\) of benzene, its vapour pressure decreases from \(650 \mathrm{~mm} \mathrm{Hg}\) to \(640 \mathrm{~mm}\) \(\mathrm{Hg}\). The depression of freezing point of Benzene (in \(\mathrm{K}\) ) upon addition of the solute is (Given data : Molar mass and the molal freezing point depression constant of benzene are \(78 \quad \mathrm{~g} \quad \mathrm{~mol}^{-1} \quad\) and \(5.12 \quad \mathrm{~K} \quad \mathrm{~kg} \quad \mathrm{~mol}^{-1}\), respectively)

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{ }^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution \(\mathrm{M}\) is (a) \(39.3 \mathrm{~mm} \mathrm{Hg}\) (b) \(36.0 \mathrm{~mm} \mathrm{Hg}\) (c) \(29.5 \mathrm{~mm} \mathrm{Hg}\) (d) \(28.8 \mathrm{~mm} \mathrm{Hg}\)

The vapour pressure of acetone at \(20^{\circ} \mathrm{C}\) is 185 torr. When \(1.2 \mathrm{~g}\) of a non-volatile substance was dissolved in \(100 \mathrm{~g}\) of acetone at \(20{ }^{\circ} \mathrm{C}\), its vapour pressure was 183 torr. The molar mass \(\left(\mathrm{g} \mathrm{mol}^{-1}\right)\) of the substance is : (a) 128 (b) 488 (c) 32 (d) 64
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.