# What is the formula of normality(solution based ques.)

What is the formula of normality(solution based ques.)

## Having known that the GEW of H2SO4 is 49 g, it means that if the entire 49g are dissolved in 1 liter of water, the normality of the solution is 1N

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xxxxxxx xxx xxxxxxxxx formula using xxxxxxxx xxxxx xxxxxxxxx

The normality xx a solution xx xxxxxxx defined as xxx gram xxxxxx xx a xxxxxxxx that xx equivalent to a xxxxxx xxxxxx of a solute xxxx xxxxxxxxx to a xxxxx of x xxxxxxxxx xxx basic xxxxxxx xx xxxxxxxxx xxx

Equivalents/volume xxxxx xxxxxxxxxxx xx xxx mass xx xxxxxx xxxxxxx xx xxxxxxxxxx xxxx xxx the equivalent xxxx is obtained xx xxxxxxxx xxx molecular xxxx xx xxx basicity xx the solution.

xxxxxxxx xx illustrate:

Example xx

xxxxxx xxxxx that xxx xxx of xxxxx xx 49 g, xx means that if xxx xxxxxx xxx are xxxxxxxxx in x liter xx xxxxxx xxx xxxxxxxxx of xxx solution xx xx

EXAMPLE xx 100 xxxxx of a xxxxxx that is xxxx xxx xxxx xx come xx xxxx a xxxxxxxxx The solute xxx x density xx xxxxxxx and x 70% essay.

xxxxxxx x 0.70 = 0.84g xxxxxxxxx

xxxx divide xx xxxxxxxxxx x 119.05 xx

xxxx means xxxx only xxxxxxxx of the xxxxxx that we xxxx xxxx xx needed.

xx xxxxxxxxxxx xxx xxxxxxxxx you xxxx be xxxxxxxx xx multiply the xxxxxx xxxxxxx xx xxx xxxxxx xx get xxx xxxxx xxxxxx

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Finding xxx normality formula xxxxx solution based xxxxxxxxx

The normality of x xxxxxxxx is xxxxxxx defined as xxx gram weight xx x xxxxxxxx xxxx xx xxxxxxxxxx to x xxxxxx xxxxxx xx x solute xxxx referring to a liter of a xxxxxxxxx The xxxxx xxxxxxx xx xxxxxxxxx is:

Equivalents/volume where equivalents xx the mass xx xxxxxx xxxxxxx by equivalent xxxx xxx the equivalent xxxx xx xxxxxxxx xx xxxxxxxx xxx molecular xxxx by the basicity xx xxx solution.

xxxxxxxx to illustrate:

Example 1:

xxxxxx known that the GEW xx xxxxx is 49 xx xx xxxxx that if the entire 49g are xxxxxxxxx in 1 xxxxx xx water, xxx xxxxxxxxx of xxx solution is xx

EXAMPLE 2: 100 xxxxx xx x solute xxxx is xxxx xxx xxxx to xxxx up with a xxxxxxxxx xxx solute has x xxxxxxx of 1.2g/ml xxx a 70% xxxxxx

1.2g/ml x xxxx x xxxxx solute/ml

100g xxxxxx xx (0.84g/ml) = xxxxxx xx

xxxx xxxxx xxxx only xxxxxxxx xx xxx xxxxxx that xx have will be xxxxxxx

xx calculating xxx xxxxxxxxx xxx xxxx xx xxxxxxxx xx multiply xxx xxxxxx xxxxxxx of xxx liquid xx get xxx xxxxx xxxxxx

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