Molarity of a 50 mL H_2SO_4 solution is 10.0 | vedclass

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(B) Given, molarity of $H_2SO_4$ solution $= 10.0 \ M$. Volume $(V) = 50 \ mL$. Density of solution $(d) = 1.4 \ g/cc$. Molarity $10 \ M$ means $10 \

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$23.8$

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(B) Given, molarity of $H_2SO_4$ solution $= 10.0 \ M$. Volume $(V) = 50 \ mL$. Density of solution $(d) = 1.4 \ g/cc$. Molarity $10 \ M$ means $10 \ moles$ of $H_2SO_4$ are present in $1000 \ mL$ of solution. Molar mass of $H_2SO_4$ $(M_B) = 98 \ g/mol$. Mass of $H_2SO_4$ in $1000 \ mL = 10 \ mol 	imes 98 \ g/mol = 980 \ g$. Mass of $H_2SO_4$ in $50 \ mL = (980 \ g / 1000 \ mL) 	imes 50 \ mL = 49 \ g$. Mass of solution $= d 	imes V = 1.4 \ g/cc 	imes 50 \ mL = 70 \ g$. Mass of solvent $(w_A) = 	ext{Mass of solution} - 	ext{Mass of solute} = 70 \ g - 49 \ g = 21 \ g$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{49 \ g / 98 \ g/mol}{21 \ g / 1000 \ g/kg} = 0.5 \ mol 	imes (1000 / 21) \ kg^{-1} \approx 23.8 \ m$.

List-$I$	List-$II$
$(a)$ $2 \text{ moles of ethene}$	$(i)$ $11.2 \text{ L volume at STP}$
$(b)$ $\text{Molar mass is equal to } 66 \text{ g}$	$(ii)$ $56 \text{ g}$
$(c)$ $1 \text{ g of } H_2$	$(iii)$ $12.04 \times 10^{23} \text{ molecules}$
$(d)$ $2 \text{ moles of water vapours}$	$(iv)$ $1.5 \text{ mole of } CO_2$

Molarity of a $50 \ mL$ $H_2SO_4$ solution is $10.0 \ M$. If the density of the solution is $1.4 \ g/cc$, calculate its molality.

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