Question #c0970

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Answer 1 · 2018-02-01T11:56:18

$1.63 "mol"$ of sodium chloride.

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

Therefore, we must first calculate the mass of $NaCl$ present.

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

The density of sodium chloride is $2.16g/(mL)$ .

So input: $2.16=m/44$

$2.16*44=m/44*44$

$m=95.04g$ of sodium chloride.

Now we may use $n=m/M$ .

The molar mass of sodium chloride is $58.4g/("mol")$ .

So simply input: $n=95.04/58.4$

$n=1.63 "mol"$ of sodium chloride.

Answer 2 · 2018-02-01T11:59:19

$~~1.63mol$

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

First, know that $1mL=1cm^3$ , so

$44mL=44cm^3$

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

$rho=m/V$ or $m=rho*V$

The density of sodium chloride is $2.16g"/"cm^3$

$:."mass"=2.16g"/"cm^3*44cm^3=95.04g$

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

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

$"moles"="mass"/"molar mass"$

The molar mass of sodium chloride is $58.4g"/"mol$ .

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

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