Question

Calculate molarity of solution containing 16 gm glucose per300 ml solution??
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Answer 1

Concept : MOLARITY

Solution :

$\boxed{\mathsf{Molarity, M\:=\:{\frac{Number\:of\:moles\:of\:solute} {Volume \:of\:the\:solution\:(\:in\:l \:)}}}}$

Given mass of Glucose, $\mathsf{C_6\:H_{12}\:O_6}$ = 16 g.

For finding Molarity, we have to calculate its number of moles of solute i. e. Glucose.

Number of moles, n = $\mathsf{\frac{Mass\:of\:Glucose } {Molar\:Mass\:of\:Glucose\: }}$

Molar Mass of Glucose, $\mathsf{C_6\:H_{12}\:O_6}$ = 12×6 + 1×12 + 16×6 = 72 + 12 + 96 = 180 g

Number of moles, n = $\mathsf{\frac{16} {180}}$ = $\mathsf{\frac{4} {45}}$

Now, We have,

Number of moles of solute, Glucose = $\mathsf{\frac{4} {45}}$

And Volume of the solution is given = 300 ml

But we have to change it's unit from 'ml' to 'l' as per the formula of Molarity.

Now,

Molarity, M = $\mathsf{\frac{Number\:of\:moles\:of\:solute} {Volume \:of\:the\:solution\:(\:in\:l \:) }}$

Molarity, M = $\mathsf{\frac{4*1000}{45*300}}$

➡️ Molarity, M = $\mathsf{\frac{8}{27}\:=\:0.296\:M}$

About Molarity :

Molarity of the solution may be defined as the number of moles of solute present in 1 litre of solution.

Formula :

$\boxed{\mathsf{Molarity, M\:=\:{\frac{Number\:of\:moles\:of\:solute} {Volume \:of\:the\:solution\:(\:in\:l \:)}}}}$

It is dependent on temperature as volume changes with the change in temperature.

It's unit is ' mole per litre' ( $\mathsf{mol\:{L} ^{-1}}$ or ' moles per cubic decimetre' ( $\mathsf{mol\:{dm} ^{-3}}$ )

It's symbol is M.