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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

When 25g of...

Question

When $25 g$ of ${N a}_{2} {S O}_{4}$ is dissolved in $10^{3} k g$ of solution, its concentration will be :

2.5 p p m

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25 p p m

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250 p p m

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100 p p m

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Solution

The correct option is B $25 p p m$
When a solute is present in very minute amounts (trace quantities) the concentration is expressed in ppm.
ppm of $A = \frac{m a s s o f c o m p o n e n t A}{t o t a l m a s s o f s o l u t i o n} \times 10^{6} = \frac{0.025}{10^{3}} \times 10^{6} = 25 p p m$

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Similar questions

Q. Assertion :Heat of solution is positive when

{N a}_{2} {S O}_{4} \cdot 10 H_{2} O

is dissolved in water but it is negative when anhydrous

C u {S O}_{4}

is dissolved in water. Reason: Molar masses of

{N a}_{2} {S O}_{4} \cdot 10 H_{2} O

and

C u {S O}_{4}

are different.

1.25 g

of a sample of

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

and

{N a}_{2} {S O}_{4}

is dissolved in

250 m l

solution.

25 m l

of this solution neutralises

20 m l

0.1 N

H_{2} {S O}_{4}

. The % of

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

in this sample is:

25 g m

mixture of

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

and

{N a}_{2} {S O}_{4}

are dissolved in

250 m l

of solution.

25 m l

of above solution required

20 m l

0.1 N

H_{2} {S O}_{4}

using phenolphthalein indicator. The mass of

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

in the mixture is:

Q. Solid

B a {({N O}_{3})}_{2}

is gradually dissolved in a

1 \times 10^{- 4} M

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

solution. At what concentration of

{B a}^{2 +}

will precipitate begin to form? (

K_{s p}

for

B a {C O}_{3} = 5.1 \times 10^{- 9}

)

100 m l

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

solution containing

5.3 g

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

was exactly neutralised by

200 m l

H_{2} {S O}_{4}

solution. What is the pH of

H_{2} {S O}_{4}

(

l o g 5 = 0.6990

)

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When 25g of Na2SO4 is dissolved in 103kg of solution, its concentration will be :

The correct option is B 25 ppmWhen a solute is present in very minute amounts (trace quantities) the concentration is expressed in ppm.ppm of A=massofcomponentAtotalmassofsolution×106=0.025103×106=25 ppm

When $25 g$ of ${N a}_{2} {S O}_{4}$ is dissolved in $10^{3} k g$ of solution, its concentration will be :

The correct option is B $25 p p m$
When a solute is present in very minute amounts (trace quantities) the concentration is expressed in ppm.
ppm of $A = \frac{m a s s o f c o m p o n e n t A}{t o t a l m a s s o f s o l u t i o n} \times 10^{6} = \frac{0.025}{10^{3}} \times 10^{6} = 25 p p m$