Calculate the molarity of each of the following: (a) $30 g$ of ${C o (N O}_{3})_{2}$ . ${6 H}_{2} O$ in $4.3 L$ of solution (b) $30 m L$ of 0.5 $M H_{2} {S O}_{4}$ diluted to $500 m L$ .

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Solution

(a) $molarity = \frac{No. of moles of solute}{{Volume of soluion}_{(in L)}}$
Given that mass of $C o {(N O_{3})}_{2} .6 H_{2} O = 30 g$
Molar mass of $C o {(N O_{3})}_{2} .6 H_{2} O = 291 g / m o l$
$∴$ No. of moles of $C o {(N O_{3})}_{2} .6 H_{2} O = \frac{30}{291} = 0.103 mol$
Volume of solution $= 4.3 L$
$∴ molarity = \frac{0.103}{4.3} = 0.04 M$
(b) By using formula, we have
$M_{1} V_{1} = M_{2} V_{2}$
Given that:-
Volume of $H_{2} S O_{4} (V_{1}) = 30 m L$
Diluted volue of $H_{2} S O_{4} (V_{2}) = 500 m L$
Final molarity $(M_{2}) = ?$
Initial molarity of $H_{2} S O_{4}$ required.
Hence the required answer for is- $0.04 M$

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Calculate the molarity of each of the following: (a) 30g of Co(NO3)2. 6H2O in 4.3L of solution (b) 30mL of 0.5 M H2SO4 diluted to 500mL.

Calculate the molarity of each of the following: (a) $30 g$ of ${C o (N O}_{3})_{2}$ . ${6 H}_{2} O$ in $4.3 L$ of solution (b) $30 m L$ of 0.5 $M H_{2} {S O}_{4}$ diluted to $500 m L$ .