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

Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

1 g sample of...

Question

$1$ g sample of washing soda was dissolved in $50$ mL of $1 M H C l$ . The excess acid required for the neutralization of $21.13$ mL of $1.018 M B a (O H_{2})$ . The degree of hydration of washing soda is :

Open in App

Solution

Meq. of $H C l$ added in washing soda $= 50 \times 1 = 50$
Meq. of $H C l$ used for $B a (O H_{2}) =$ Meq. of $B a (O H_{2})$ $= 21.13 \times 1.018 \times 2 = 43.02$
Meq. of $H C l$ used for washing soda $= 50 - 43.02 = 6.98$
Meq. of washing soda $({N a}_{2} C O_{3}) \cdot x H_{2} O = 6.98$
or, $\frac{w}{\frac{M}{2}} \times 1000 = 6.98$ $⟹ \frac{1 \times 2}{M} \times 1000 = 6.98 (∵ w = 1 g)$
Therefore, $M = 286.5 g {m o l}^{- 1}$
Thus, molar mass of $({N a}_{2} C O_{3}) \cdot x H_{2} O$ $= 46 + 60 + 18 x = 106 + 18 x$ $106 + 18 x = 286.5 o r x = 10$
Therefore, the degree of hydration $= 10$

Suggest Corrections

Similar questions

1.64

g of a mixture of

C a C O_{3}

and

M g C O_{3}

was dissolved in

50

mL of

0.8 M H C l

. The excess of acid required

16

mL of

0.25 M N a O H

for neutralization. Calculate the percentage of

C a C O_{3}

in the sample.

Q. Washing soda

(N a_{2} C O_{3} .10 H_{2} O)

is widely used in softening of hard water. If

1

L of hard water requires

0.0286

g of washing soda, the hardness of

C a C O_{3}

in ppm is :

1.20 g

sample of

N a_{2} C O_{3}

and

K_{2} C O_{3}

was dissolved in water to form

100 m L

of a solution.

20 m L

of this solution required

40 m L

0.1 N H C l

for complete neutralization. Calculate the weight of precipitate in the mixture if another

20 m L

of this solution is treated with excess of

B a C l_{2}

Q. A

1.5

g sample containing oxalic acid and some inert impurity was dissolved in enough water and volume was made up to

250

mL. A

20

mL portion of this solution was then mixed with

30

mL of an alkali solution. The resulting solution was then treated with stoichiometric amount of

C a C l_{2}

just needed for precipitation of oxalate as

C a C_{2} O_{4}

solution was filtered off and filtrate was finally titrated against

0.1 M H C l

solution.

8.0

mL of acid was required to reach the equivalence point. At last the above neutral solution was treated with excess of

A g N O_{3}

solution and

A g C l

obtained was washed, dried and weighed to be

0.4305

g. Determine the mass percentage of oxalic acid in the original sample.

Q. Washing soda

(N a_{2} C O_{3} .10 H_{2} O)

is widely used in softening of hard water. If 1L of hard water requires 0.0143 g of washing soda, what is hardness of water in terms of ppm of

C a C O_{3}

Join BYJU'S Learning Program

1 g sample of washing soda was dissolved in 50 mL of 1MHCl. The excess acid required for the neutralization of 21.13 mL of 1.018M Ba(OH2). The degree of hydration of washing soda is :

$1$ g sample of washing soda was dissolved in $50$ mL of $1 M H C l$ . The excess acid required for the neutralization of $21.13$ mL of $1.018 M B a (O H_{2})$ . The degree of hydration of washing soda is :