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Equivalent Mass

A sample of 0...

Question

A sample of $0.318 g$ of pure sodium carbonate is titrated with a $H C l$ solution. A volume of $60 mL$ is required to reach the end point. Calculate the molarity of the acid.

0.1 M

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0.4 M

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0.2 M

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None of these

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Solution

The correct option is A $0.1 M$
n-factor of $N a_{2} C O_{3} = 2$ , n-factor for $H C l = 1$
At the equivalence point,
Equivalents of $N a_{2} C O_{3}$ = equivalents of $H C l$
$n-factor \times \frac{weight of N a_{2} C O_{3}}{molar mass} = molarity \times volume in L$
$2 \times \frac{0.318}{106} = M \times 0.06$
On solving, the molarity of $H C l$ $= 0.1 M$

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Similar questions

Q. A sample of

0.318 g

of pure sodium carbonate is titrated with

H C l

solution. A volume of

60 mL

is required to reach the end point. Calculate the molarity of the acid.

Q. A sample of

0.636 g

of pure sodium carbonate is titrated with

H C l

solution. A volume of

60 mL

is required to reach the end point. Calculate the molarity of the acid.

20 g

of a sample of

B a (O H)_{2}

is dissolved in

10 m L

0.5 N H C l

solution. The excess of

H C l

was titrated with

0.2 N N a O H .

The volume of

N a O H

required was

10 m L

. Calculate the percentage of

B a (O H)_{2}

in the sample.
(Given molar mass of

B a

137 g m o l^{- 1}

)

Q. A

40.0 m L

sample of

0.100 M H N O_{3}

is titrated with

0.200 M N a O H

.
Calculate the volume required to reach the equivalence point.

Q. A

20 m L

sample of a

B a (O H)_{2}

solution is titrated with

0.245 M H C l

. If

27.15 m L

H C l

is required, what is the molarity of the

B a (O H)_{2}

solution?

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Equivalent Mass

Standard XIII Chemistry

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A sample of 0.318 g of pure sodium carbonate is titrated with a HCl solution. A volume of 60 mL is required to reach the end point. Calculate the molarity of the acid.

The correct option is A 0.1 Mn-factor of Na2CO3=2, n-factor for HCl=1 At the equivalence point, Equivalents of Na2CO3 = equivalents of HCl n-factor×weight of Na2CO3molar mass=molarity×volume in L 2×0.318106=M×0.06 On solving, the molarity of HCl =0.1 M

A sample of $0.318 g$ of pure sodium carbonate is titrated with a $H C l$ solution. A volume of $60 mL$ is required to reach the end point. Calculate the molarity of the acid.