10.6 , g of a substance of molecular weight 1 | vedclass

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(B) Step $1$: Calculate the initial molarity of the solution.Initial Molarity $(M_1) = \frac{	ext{mass}}{	ext{molecular weight}} 	imes \frac{1000}{

Vedclass · Accepted Answer

$10^{-2} \, M$

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(B) Step $1$: Calculate the initial molarity of the solution.Initial Molarity $(M_1) = \frac{	ext{mass}}{	ext{molecular weight}} 	imes \frac{1000}{V \, (mL)}$$M_1 = \frac{10.6}{106} 	imes \frac{1000}{100} = 0.1 	imes 10 = 1 \, M$Step $2$: Use the dilution formula $M_1 V_1 = M_2 V_2$ for the dilution process.Here,$M_1 = 1 \, M$,$V_1 = 10 \, mL$,and $V_2 = 1000 \, mL$.$1 	imes 10 = M_2 	imes 1000$$M_2 = \frac{10}{1000} = 0.01 \, M = 10^{-2} \, M$Therefore,the correct option is $B$.

$10.6 \, g$ of a substance of molecular weight $106$ was dissolved in $100 \, mL$. $10 \, mL$ of this solution was pipetted out into a $1000 \, mL$ flask and made up to the mark with distilled water. The molarity of the resulting solution is

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