Home
Class 12
CHEMISTRY
A bottle, which contains 200 ml of 0.1 M...

A bottle, which contains 200 ml of 0.1 M KOH, absorbs 1 millimole of `CO_(2)` from the air. If the solution is then treated with standard acid using phenolphthalein indicator, the normality of solution formed by absorption of `CO_(2)` is :

A

0.095 N

B

0.1 N

C

0.2 N

D

0.036 N

Text Solution

Verified by Experts

The correct Answer is:
A

milli equivalent of acid = milli equivalent of `K_2 CO_3` + milli equivalent of KOH =1+8 = 19
`N= ("mill equivalent")/(V(ml)) = (19)/(200) = 0.095 N` .
Promotional Banner

Topper's Solved these Questions

  • MOCK TEST 5

    VMC MODULES ENGLISH|Exercise CHEMISTRY (SECTION 2)|5 Videos

  • MOCK TEST 4

    VMC MODULES ENGLISH|Exercise CHEMISTY (SECTION 2)|5 Videos

  • MOCK TEST 6

    VMC MODULES ENGLISH|Exercise CHEMISTRY (SECTION 2)|5 Videos

Similar Questions

Explore conceptually related problems

Calculate the pH of a solution which contains 100mL of 0.1 M HC1 and 9.9 mL of 1.0 M NaOH .

If 200 ml of 0.1 MH_2SO_4 solution is mixed with 200 ml of 0.15 M NaOH solution then heat evolve during the neutralization process is

The weight of solute present in 200mL of 0.1M H_(2)SO_(4) :

Air sample from an industrial area of Delhi, which is heavily polluted by CO_(2) , was collected and analysed. One such sample of 224 L of air measured at STP was passed through a 500 " mL of " 0.1 M KOH solution, where CO_(2)(g) was absorbed completely. 50 " mL of " the above solution was then treated with excess of BaCl_(2) solution where all the carbonate was precipitated as BaCO_(4)(s) . The solution was filtered off and the filtrate required 30 " mL of " 0.1 M HCl solution for neutralisation. Q. ppm strength of CO_(2)(g) volume by volume (" mL of " CO_(2) per 10^(6) L of air) is

Air sample from an industrial area of Delhi, which is heavily polluted by CO_(2) , was collected and analysed. One such sample of 224 L of air measured at STP was passed through a 500 " mL of " 0.1 M KOH solution, where CO_(2)(g) was absorbed completely. 50 " mL of " the above solution was then treated with excess of BaCl_(2) solution where all the carbonate was precipitated as BaCO_(4)(s) . The solution was filtered off and the filtrate required 30 " mL of " 0.1 M HCl solution for neutralisation. Q. Calculate the weight of the precipitate of BaCO_(3)(s) obtained from 50 " mL of " the above test solution. (Ba=137,C=12,O=16,Mw(BaCO_(3))=137+12+3xx16=197gmol^(-1))

1.7225 g of a metal (bivalent) salt A_(x)(CO_3)_(y)(OH)_(z) was dissolved in water to make 100 " mL of " solution 50 " mL of " this solution required 10 " mL of " 0.5 M H_2SO_4 solution for complete neutralisation using phenolphthalein indicator. Another 50 mL solution required 15 " mL of " same acid using methyl orange indicator. Deduce the formula of the salt.

100 ml of 1M NaOH is mixed with 50ml of 1N KOH solution. Normality of mixture is

Calculate the normality of a solution obtained by mixing 200 mL of 1.0 N NaOH and 100 mL of pure water.

A solution containing 200 ml 0.5 M KCl is mixed with 50 ml 19% w/v MgCl_(2) and resulting solution is dilute 8 times. Molarity of chloride ion is final solution is :

0.1g of a solution containing Na_(2)CO_(3) and NaHCO_(3) requires 10mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator, mass% of Na_(2)CO_(3) in solution is: