Chapter 3, Problem 3.65PAE

3.63 How many moles of solute are present in each of these

solutions?

(a) 48.0 mL of 3.4 M H₂SO₄.

(b) 1.43 mL of 5.8 M KNO₃.

(c) 321 L of 0.034M NH₃

(d) 1.9 × 10-3 L of 1.4 × 10-5 M NaF

Interpretation Introduction

To determine:

The number of moles of solute for 48.0 mL of 3.4 M $H_{2}S{O}_4$ solution.

Explanation:

Molarity is defined as the number of moles of solute ( $H_{2}S{O}_4$ ) in 1 L of solution.

The solution is 3.4 M, i.e. it contains 3.4 moles of $H_{2}S{O}_4$

per 1000 mL solution. Thus, for 48.0 mL solution, the number of moles would be:

$n=3.4\times \frac{48.0mL}{1000mL}=0.16\text{}moles$

Explanation of Solution

Molarity is defined as the number of moles of solute ( $H_{2}S{O}_4$ ) in 1 L of solution.

The solution is 3.4 M, i.e. it contains 3.4 moles of $H_{2}S{O}_4$

per 1000 mL solution. Thus, for 48.0 mL solution, the number of moles would be:

$n=3.4\times \frac{48.0mL}{1000mL}=0.16\text{}moles$

Conclusion

Therefore, knowing the molarity of a given solution, the number of moles can be calculated for any given volume by simple calculations.

Interpretation Introduction

To determine:

The number of moles of solute for 1.43 mL of 5.8 M $KN{O}_3$ solution.

Explanation:

Molarity is defined as the number of moles of solute ( $KN{O}_3$ ) in 1 L of solution.

The solution is 5.8 M, i.e. it contains 5.8 moles of $KN{O}_3$

per 1000 mL solution. Thus, for 1.43 mL solution, the number of moles would be:

$n=5.8\times \frac{1.43mL}{1000mL}=0.0083\text{}moles$

Explanation of Solution

Molarity is defined as the number of moles of solute ( $KN{O}_3$ ) in 1 L of solution.

The solution is 5.8 M, i.e. it contains 5.8 moles of $KN{O}_3$

per 1000 mL solution. Thus, for 1.43 mL solution, the number of moles would be:

$n=5.8\times \frac{1.43mL}{1000mL}=0.0083\text{}moles$

Conclusion

Therefore, knowing the molarity of a given solution, the number of moles can be calculated for any given volume by simple calculations.

Interpretation Introduction

To determine:

The number of moles of solute for 3.21 L of 0.034 M $N{H}_3$ solution.

Explanation:

Molarity is defined as the number of moles of solute ( $N{H}_3$ ) in 1 L of solution.

The solution is 0.034 M, i.e. it contains 0.034 moles of $N{H}_3$

per 1000 mL solution. Thus, for 3.21 L solution, the number of moles would be:

$n=0.034\times \frac{3.21L}{1L}=0.11\text{}moles$

Explanation of Solution

Molarity is defined as the number of moles of solute ( $N{H}_3$ ) in 1 L of solution.

The solution is 0.034 M, i.e. it contains 0.034 moles of $N{H}_3$

per 1000 mL solution. Thus, for 3.21 L solution, the number of moles would be:

$n=0.034\times \frac{3.21L}{1L}=0.11\text{}moles$

Conclusion

Therefore, knowing the molarity of a given solution, the number of moles can be calculated for any given volume by simple calculations.

Interpretation Introduction

To determine:

The number of moles of solute for $1.9\times {10}^{3}L$ of $1.4\times {10}^{-5}M$

$NaF$ solution.

Explanation:

Molarity is defined as the number of moles of solute ( $NaF$ ) in 1 L of solution.

The solution is $1.4\times {10}^{-5}M$, i.e. it contains $1.4\times {10}^{-5}$ moles of $NaF$ per 1000 mL solution. Thus, for $1.9\times {10}^{3}L$

solution, the number of moles would be:

$n=1.4\times {10}^{-5}\times \frac{1.9\times {10}^{3}L}{1L}=0.027\text{}moles$

Explanation of Solution

Molarity is defined as the number of moles of solute ( $NaF$ ) in 1 L of solution.

The solution is $1.4\times {10}^{-5}M$, i.e. it contains $1.4\times {10}^{-5}$ moles of $NaF$ per 1000 mL solution. Thus, for $1.9\times {10}^{3}L$

solution, the number of moles would be:

$n=1.4\times {10}^{-5}\times \frac{1.9\times {10}^{3}L}{1L}=0.027\text{}moles$

Conclusion

Therefore, knowing the molarity of a given solution, the number of moles can be calculated for any given volume by simple calculations.