Question

How many liters of a $2.50{\mathrm{MKNO}}_3$solution can be prepared from$60.0\mathrm{g}$of ${\mathrm{KNO}}_3$ ?

Short answer

$0.237L$of a $2.50{\mathrm{MKNO}}_3$ solution can be prepared from$60.0g$of${\mathrm{KNO}}_3$.

Step-by-step solution

Given information

To calculate the how much liters of a $2.50{\mathrm{MKNO}}_3$solution can be prepared from$60.0g$of${\mathrm{KNO}}_3$.

To calculate the moles of KNO3 by using the following expression 

Moles of ${\mathrm{KNO}}_3=\frac{\text{mass of}{\mathrm{KNO}}_3}{\text{molar mass of}{\mathrm{KNO}}_3}$

$=\frac{60\mathrm{g}}{101.1\mathrm{g}/\mathrm{mol}}$

$=0.593\mathrm{mol}$.

To calculate the volume required by using the following expression :

Volume of ${\mathrm{KNO}}_3=\frac{\text{Moles of}{\mathrm{KNO}}_3}{\text{Molarity of}{\mathrm{KNO}}_3}$

$=\frac{0.593\mathrm{mol}}{2.50\mathrm{M}}$

$=0.237\mathrm{L}$

Therefore, the required volume of${\mathrm{KNO}}_3$solution is$0.237L$.