Question

50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol

Answer 1

Answer:

Approximately $0.47\; \rm mol \cdot L^{-1}$ (note that $1\; \rm M = 1 \; \rm mol \cdot L^{-1}$.)

Explanation:

The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this $\rm NaNO_3$ solution in water,

Let $n$ be the number of moles of the solute in the whole solution. Let $V$ represent the volume of that solution. The formula for the molarity $c$ of that solution is:

$\displaystyle c = \frac{n}{V}$.

In this question, the volume of the solution is known to be $1250\; \rm mL$. That's $1.250\; \rm L$ in standard units. What needs to be found is $n$, the number of moles of $\rm NaNO_3$ in that solution.

The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of $\rm NaNO_3$ is $85\; \rm g \cdot mol^{-1}$ means that the mass of one mole of

$\displaystyle n = \frac{m}{M}$.

For this question,

$\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}$.

Calculate the molarity of this solution:

$\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}$.

Note that $1\; \rm mol \cdot L^{-1}$ (one mole per liter solution) is the same as $1\; \rm M$.