Chapter 9: Q. 9.101 (page 318)

How many milliliters of a $1.75 M L i C l$ solution contain $15.2 g$ of $L i C l$ ?

Short Answer

Expert verified

The solution is $204 m l L i C l$ of $1.75 M L i C l$ solution contains $15.2 g$ of $L i C l$ .

Step by step solution

Given information

To find how many milliliters of $1.75 M L i C l$ solution contains $15.2 g$ of $L i C l$ :

Solution for given data:

Gram $L i C l$ Molar mass moles $L i C l$ Molarity $L i C l$ metric system $m l L i C l$ .

Equalities /Conversion factors:

$1 m o l e L i C l$ role="math" localid="1652512944024" $= 42.5 g LiCl$

$\frac{1 moleLiCl}{42.5 gLiCl}$ and $\frac{42.5 gLiCl}{1 moleLiCl}$

The molarity of any solution can be written as two conversion factors

role="math" localid="1652513781092" $\begin{array}{cll} 1 LLiCl & = 1.75 m o l e s L i C l \\ \frac{1 LLiCl}{1.75 moleLiCl} & and \frac{1.75 moleLiCl}{1 LLiCl} \end{array}$

The liter of solution can be converted into milliliter of solution using the metric system

$1000 m l LiCl = 1 L L i C l$

$\frac{1000 mlLiCl}{1 LLiCl}$ and $\frac{1 LLiCl}{1000 mlLiCl}$

Final calculation :

$ml$ of $LiCl = 15.2 g LiCl 脳 \frac{1 moleLiCl}{42.5 g LiCl} 脳 \frac{1 LiCl}{1.75 mole L iCl} 脳 \frac{1000 ml LiCl}{1 L LiCl}$

$= 204 m l L i C l .$

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How many milliliters of a $1.75 M L i C l$ solution contain $15.2 g$ of $L i C l$ ?

Short Answer

Step by step solution

Given information

Solution for given data:

Equalities /Conversion factors:

Final calculation :

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91影视

How many milliliters of a1.75MLiClsolution contain 15.2gofLiCl?

Short Answer

Step by step solution

Given information

Solution for given data:

Equalities /Conversion factors:

Final calculation :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

The Earths Atmosphere

Kinetics

Chemical Reactions

Nuclear Chemistry

Organic Chemistry

Study anywhere. Anytime. Across all devices.

How many milliliters of a $1.75 M L i C l$ solution contain $15.2 g$ of $L i C l$ ?