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Submitted by kicude on Tue, 2012-07-03 01:42
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# What is the formula of normality(solution based ques.)

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

Submitted by mbitheh on Tue, 2012-07-03 06:25
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## 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 the xxxxxxxxx formula xxxxx xxxxxxxx xxxxx xxxxxxxxx

The xxxxxxxxx xx x xxxxxxxx is xxxxxxx defined xx xxx gram weight xx a xxxxxxxx xxxx is equivalent to x single weight of a xxxxxx xxxx xxxxxxxxx to x liter xx x solution. xxx basic xxxxxxx xx normality is:

Equivalents/volume xxxxx equivalents xx xxx xxxx of xxxxxx divided by equivalent xxxx xxx xxx equivalent mass is obtained xx dividing xxx xxxxxxxxx xxxx xx xxx basicity of xxx solution.

xxxxxxxx xx illustrate:

xxxxxxx 1:

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

xxxxxxx 2: xxx grams xx x xxxxxx that xx pure are used xx come xx xxxx a solution. The xxxxxx has a density of 1.2g/ml and a xxx essay.

1.2g/ml X 0.70 x xxxxx solute/ml

100g divide xx (0.84g/ml) = 119.05 ml

This xxxxx that xxxx 119.05ml xx xxx xxxxxx that we have xxxx be xxxxxxx

In xxxxxxxxxxx the normality xxx will xx xxxxxxxx xx xxxxxxxx xxx volume xxxxxxx xx the xxxxxx xx get xxx added grams.

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Finding xxx normality formula using solution based questions

xxx xxxxxxxxx xx x xxxxxxxx is xxxxxxx xxxxxxx xx the xxxx xxxxxx xx x solution xxxx xx equivalent to a xxxxxx xxxxxx of a xxxxxx when referring to x liter of x xxxxxxxxx xxx basic xxxxxxx xx normality xxx

xxxxxxxxxxxxxxxxxx where equivalents is the xxxx xx xxxxxx xxxxxxx xx equivalent xxxx xxx xxx equivalent xxxx is xxxxxxxx xx dividing the molecular xxxx by xxx xxxxxxxx xx xxx xxxxxxxxx

xxxxxxxx to illustrate:

Example 1:

xxxxxx known that the xxx xx xxxxx xx xx xx it xxxxx that xx the xxxxxx 49g xxx xxxxxxxxx xx 1 xxxxx xx xxxxxx the normality xx the xxxxxxxx xx xx

xxxxxxx xx xxx xxxxx of x xxxxxx xxxx xx pure are used to come xx with a xxxxxxxxx The solute xxx x xxxxxxx xx 1.2g/ml and a xxx essay.

xxxxxxx X xxxx x 0.84g solute/ml

xxxx divide xx xxxxxxxxxx x xxxxxx ml

This means that only 119.05ml of xxx solute that xx have will xx needed.

xx xxxxxxxxxxx the xxxxxxxxx xxx xxxx be xxxxxxxx xx xxxxxxxx the volume xxxxxxx of xxx liquid to get xxx added grams.