You visited us 1 times! Enjoying our articles? Unlock Full Access!

Ideal Gas Law

1.75 g of sol...

Question

1.75 g of solid NaOH is added to 0.25 $d m^{3}$ of 0.1 M $N i C l_{2}$ solution. Calculate mass in grams of $N i (O H)_{2}$ forms (as nearest integer).

Open in App

Solution

$2 N a O H + N i C l_{2} ⟶ N i (O H)_{2} + 2 N a C I$
Mole before reaction $\frac{1.75}{40}$ 0.25 x 0.1
$= 0.0438$ $= 0.025$
Mole left 0 $(0.025 \frac{0.0438}{2}) \frac{0.0438}{2}$ 0.0438
$∴$ Mole of $N i (O H)_{2}$ formed $= \frac{0.0438}{2}$
$∴$ wt. of $N i (O H)_{2}$ solution $= \frac{0.0438}{2} \times 92.6 = 2.0279 g$

Suggest Corrections

Similar questions

d m^{3}

N i C l_{2}

K_{s p}

N i (O H)_{2} = 1.6 \times 10^{- 14}

50 m L

0.1 M N a O H

50 m L

0.1 M

(K w = 1.00 \times 10^{- 14})

p H

10^{- 10} M

N a O H

500 m L

0.2 M

500 m L

0.2 M H C I

25^{o} C

6 g

N a O H

K_{a}

1.75 \times 10^{- 5} m o l L^{- 1}

0.30 M C a C l_{2}

400

0.20 M K C l

0.25 M C l^{-}

Join BYJU'S Learning Program