Question

The molarity of absolution of sodium chloride in 500 ml of water containing 5.85 gm of sodium chloride is

Answer 1

$\Large\underline\mathtt\orange{answer}$

Molarity =  $\tt {\frac{Number\: of\: moles \:in\: solute}{Volume \:of\: solution\: in\: litres} }$

Here, the solute is NaCl (Sodium Chloride)

Now we have to find the number of moles of NaCl

moles of NaCl =  $\tt {\frac{given\: mass}{actual\: molar\:mass\: of \:NaCl} }$

As we are given the volume of solution in milliliter, so we have to convert it into litres, so,

$\tt {500\:ml =0.5 \: litres }$

Now to find,

molarity =  $\tt \frac{number \:of\:moles\:in\:solute}{volume\:of \:solutions\:in \:litres}$

             =  $\tt \frac{0.1}{0.5}$

             =  $\tt {0.2\: moles}$

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