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Chapter 9: Q. 9.47 (page 306)

Calculate the grams of solute needed to prepare each of the following:
a. $2.00 L of a 1.50 M NaOH$ solution
b. $4.00 L of a 0.200 M KCl$ solution
c. $25.0 L of a 6.00 M HCl$ solution

Short Answer

Expert verified

a. The grams of $2.00 L of a 1.50 M NaOH$ is $120 g$ .

b. The grams of $4.00 L of a 0.200 M KCl$ is $74.55 g$ .

c. The grams of $25.0 L of a 6.00 M HCl$ is $5.47 g$ .

Step by step solution

01

Part (a) step 1: Given Information

We need to calculate the grams of $2.00 L of a 1.50 M NaOH$ solution.

02

Part (a) step 2: Explanation

Considering

$V (NaOH) = 2.00 L M = 1.5 M = 1.5 \frac{mole}{L}$

We know $n = M 脳 V$

Now, n = $1.5 脳 2 = 3 mole$

Number of moles = $\frac{mass}{molarmass}$

$m (NaOH) = n (NaOH) 脳 M (NaOH) = 3.0 mole 脳 (23 + 1 + 6) \frac{g}{mol} = 120 g$

03

Part (b) step 1: Given Information

We need to calculate the grams of $4.00 L of a 0.200 M KCl$ solution.

04

Part (b) step 2: Explanation

Considering,

$V (KCl) = 4.00 L M = 0.200 M = 0.200 \frac{mole}{L}$

we know $n = M 脳 V$

Now,

$n (K C l) = M 脳 V (K C l) = 0.200 \frac{mole}{L} 脳 4.00 L = 0.800 mole$

The number of moles = $\frac{mass}{molarmass}$

$m (KCl) = n (KCl) 脳 M (KCl) = 0.800 mole 脳 (39.10 + 35.45) \frac{g}{mol} = 74.55 g$

05

Part (c) step 1: Given Information

We need to calculate the grams of $25.0 L of a 6.00 M HCl$ solution.

06

Part (c) step 2: Explanation

Considering,

V = $25 L$ HCL

M = $6 M$

We know $n = M 脳 V$

Now,

$n (HCl) = M 脳 V (HCl) = 6.00 \frac{mole}{L} 脳 0.25 L = 0.15 mole$

The number of moles = $\frac{mass}{molarmass}$

$m (HCl) = n (HCl) 脳 M (HCl) = 0.15 mole 脳 36.45 \frac{g}{mol} = 5.47 g$

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