# Question c0970

Feb 1, 2018

#### Answer:

$1.63 \text{mol}$ of sodium chloride.

#### Explanation:

Remember that the formula for moles, $n$, is $n = \frac{m}{M}$, where $m$ is the mass of the substance, and $M$ its molar mass.

Therefore, we must first calculate the mass of $N a C l$ present.

To do this, we can use the formula for density, $\rho$, when $\rho = \frac{m}{V}$, where $V$ is the volume of liquid.

The density of sodium chloride is $2.16 \frac{g}{m L}$.

So input: $2.16 = \frac{m}{44}$

$2.16 \cdot 44 = \frac{m}{44} \cdot 44$

$m = 95.04 g$ of sodium chloride.

Now we may use $n = \frac{m}{M}$.

The molar mass of sodium chloride is $58.4 \frac{g}{\text{mol}}$.

So simply input: $n = \frac{95.04}{58.4}$

$n = 1.63 \text{mol}$ of sodium chloride.

Feb 1, 2018

#### Answer:

$\approx 1.63 m o l$

#### Explanation:

Well, we'll need a few steps here...

First, know that $1 m L = 1 c {m}^{3}$, so

$44 m L = 44 c {m}^{3}$

Now, we can use the density equation to find the mass of the sodium chloride:

$\rho = \frac{m}{V}$ or $m = \rho \cdot V$

The density of sodium chloride is $2.16 g \text{/} c {m}^{3}$

$\therefore \text{mass"=2.16g"/} c {m}^{3} \cdot 44 c {m}^{3} = 95.04 g$

So, we have $95.04 g$ of sodium chloride.

Next, we have to use the molar mass formula to find the amount of moles.

$\text{moles"="mass"/"molar mass}$

The molar mass of sodium chloride is $58.4 g \text{/} m o l$.

:."moles"=(95.04cancelg)/(58.4cancelg"/"mol)~~1.63mol#

So, our final answer is $1.63$ moles of sodium chloride.