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Molar Mass

4.0 g of caus...

Question

$4.0 g$ of caustic soda $(m o l m a s s 40)$ contains same number of sodium ions as present in ?

10.6 g

{N a}_{2} {C O}_{3} (m o l / m a s s 106)

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58.5 g

of NaCl

(f o r m u l a m a s s 58.5)

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100 m l

0.5 M {N a}_{2} {S O}_{4} (f o r m u l a m a s s 142)

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1 m o l

{N a N O}_{3} (m o l . m a s s 85)

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Solution

The correct option is A $10.6 g$ of ${N a}_{2} {C O}_{3} (m o l / m a s s 106)$
Number of mole of sodium $i o n (N a^{+}) = \frac{4}{40} = 0.1 m o l e$
(A) Mole of sodium $i o n (N a^{+}) = \frac{10.6}{100} \times 2 = 0.2 m o l e$
(B) mole of sodium $i o n (N a^{+}) = \frac{58.5}{58.5} = 1 m o l e$
(C) mole of sodium $i o n (N a^{+}) = 2 \times 100 m p \times 0.5 \times 10^{- 3} = 0.1$
(D) mole of $N a^{+} = 1 m o l e$
Option $C$ is correct option

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Similar questions

Q. A mixture of

0.84 g

N a H {C O}_{3}

4.0 g

N a O H

and

10.6 g

{N a}_{2} {C O}_{3}

is dissolved in

100 m l

H_{2} O

. The solution obtained is thus titrated with

0.5 M

H C l

solution. Which of the following option is/are correct?

Question 46

Fill in the blanks to make the statements true.

10g of caustic soda dissolved in 100ml of water makes a solution of caustic soda. Amount of caustic soda needed for 1 litre of water to make the same type of solution is ___.

Q. How many

{N a}^{+}

ions are present in

100 m L

0.25 M

N a C l

solution?

Q. A

1

molal solution of

N a C l

results when

58.5 g

of sodium chloride is dissolved in:

100 m L

solution consists

4 g

caustic soda. The normality of the solution is

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4.0 g of caustic soda (mol mass 40) contains same number of sodium ions as present in ?

The correct option is A 10.6g of Na2CO3 (mol/mass106)Number of mole of sodium ion(Na+)=440=0.1 mole(A) Mole of sodium ion(Na+)=10.6100×2=0.2 mole(B) mole of sodium ion(Na+)=58.558.5=1 mole(C) mole of sodium ion(Na+)=2×100 mp×0.5×10−3=0.1(D) mole of Na+=1 moleOption C is correct option

$4.0 g$ of caustic soda $(m o l m a s s 40)$ contains same number of sodium ions as present in ?