Calculate the molarity of a solution of 30 g of Co(NO3)2.6H2O in 4.3 L of solution ? Consider atomic mass of Co = 59 u , N = 14 u , O = 16 u , H = 1 u

Question

Calculate the molarity of a solution of 30 g of Co(NO3)2.6H2O in 4.3 L of solution ? Consider atomic mass of Co = 59 u , N = 14 u , O = 16 u , H = 1 u  (A)  0.023 M  (B)  0.23 M  (C)  0.046 M  (D)  0.46 M

Tardigrade Admin · Accepted Answer

Mass = 30 g Molar mass =59+14× 2+12× 16+12×1=291 Moles of Co(NO3)2⋅ 6H2O=(30/291) Molarity =(30/291 × 4.3) =0.023 M

Q. Calculate the molarity of a solution of $30 g$ of $C o (N O_{3})_{2} .6 H_{2} O$ in $4.3 L$ of solution ? Consider atomic mass of $C o = 59 u$ , $N = 14 u$ , $O = 16 u$ , $H = 1 u$

$0.023 M$

$0.23 M$

$0.046 M$

$0.46 M$

Mass = $30 g$
Molar mass $= 59 + 14 \times 2 + 12 \times 16 + 12 \times 1 = 291$
Moles of $C o (N O_{3})_{2} \cdot 6 H_{2} O = \frac{30}{291}$
Molarity $= \frac{30}{291 \times 4.3}$
$= 0.023 M$

Q. Calculate the molarity of a solution of 30g of Co(NO3​)2​.6H2​O in 4.3L of solution ? Consider atomic mass of Co=59u , N=14u , O=16u , H=1u

0.023M

0.23M

0.046M

0.46M