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 \

Vedclass · Accepted Answer

$23.8$

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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 $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{49 \ g / 98 \ g/mol}{21 \ g / 1000 \ g/kg} = 0.5 \ mol 	imes (1000 / 21) \ kg^{-1} \approx 23.8 \ m$.

Column - $A$	Column - $B$
$(A)$ $88 \ g$ of $CO_2$	$(1)$ $0.2 \ mol$
$(B)$ $6.022 \times 10^{23}$ molecules of water	$(2)$ $2 \ mol$
$(C)$ $5.6 \ L$ of $O_2$ at $STP$	$(3)$ $1 \ mol$
$(D)$ $96 \ g$ of $O_2$	$(4)$ $6.022 \times 10^{23}$ molecules
$(E)$ One mole of any gas at $STP$	$(5)$ $3 \ mol$

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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