`0.80g` is impure `(NH_4)_2SO_4` was boiled with `100mL` of `0.2N NaOH` solution till all the `NH_3 (g)` evolved. The remaining solution was diluted to `250mL. 25mL` of this solution was neutralized using `5mL` of `0.2 NH_2 SO_4` solution. The percentage purity of the `(NH_4)_2SO_4` sample is :

0.80g of impure (NH_(4))_(2) SO_(4) was boiled with 100mL of a 0.2N NaOH solution till all the NH_3 (g) evolved. the remaining solution was diluted to 250 mL . 25 mL of this solution was neutralized using 5mL of a 0.2N H_(2)SO_(4) solution. The percentage purity of the (NH_(4))_(2)SO_(4) sample is:

0.7g of (NH_(4))_(2)SO_(4) sample was boiled with 100mL of 0.2 N NaOH solution was diluted to 250 ml. 25mL of this solution was neutralised using 10mL of a 0.1 N H_(2)SO_(4) solution. The percentage purity of the (NH_(4))_(2)SO_(4) sample is :

5g sample contain only Na_2CO_3 and Na_2SO_4 . This sample is dissolved and the volume made up to 250mL. 25mL of this solution neutralizes 20mL of 0.1M H_2SO_4 . Calculate the percent of Na_2SO_4 in the sample.

20mL of 0.2MNaOH(aq) solution is mixed with 35mL of 0.1MNaOH(aq) solution and the resultant solution is diluted to 100mL. 40mL of this diluted solution reacted with 10% impure sample of oxalic acid (H_2C_2O_4) . The mass impure sample is :

If 100 mL of 1 N H_(2)SO_(4) is mixed with 100 mL of 1 M NaOH solution. The resulting solution will be

5 g sample contain only Na_(2)CO_(3) and Na_(2)SO_(4) . This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M H_(2)SO_(4) . Calcalute the % of Na_(2)SO_(4) in the sample .

100 ml of 0.1 N NaOH is mixed with 50 ml of 0.1 N H_2SO_4 . The pH of the resulting solution is

Volume of 0.1 M H_2SO_4 solution required to neutralize 20 ml of 0.2 M NaOH solution is :

Volume of 0.1 M H_2SO_4 solution required to neutralize 40 ml of 0.2 M NaOH solution is :

Volume of 0.1 M H_2SO_4 solution required to neutralize 10 ml of 0.2 M NaOH solution is :