I'm having difficulty in solving the following Chemistry questions.
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I'm having difficulty in solving the following Chemistry questions.

[From: ] [author: ] [Date: 11-08-07] [Hit: ]
A 4.9%(w/w) solution of H2SO4 has a density of 1.294g/mL. Calculate its molarity.-You must thoroughly learn the concept of molarity, molality and how specific gravity (or density) comes into the picture.......
If anyone knows how to solve them, then, Plz help me out...!

1.Calculate the molarity of the following solutions:-
(a) 4g of caustic soda is dissolved in 200mL of the solution.
(b) 5.3g of anhydrous sodium carbonate is dissolved in 100mL of solution.

2. The density of a solution containing 13% by mass of sulphuric acid is 1.09g/mL. Calculate the molarity & normality of the solution.

3.2 metallic oxides contain 27.6% & 30% oxygen in them respectively. If the formula of the 1st oxide is M3O4, then, what will be that of the 2nd?

4. Insulin contains 3.4% sulphur. Calculate minimum molecular mass of insulin.

5. A 4.9%(w/w) solution of H2SO4 has a density of 1.294g/mL. Calculate its molarity.

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You must thoroughly learn the concept of molarity, molality and how specific gravity (or density) comes into the picture. Once you get to know the concepts, everything becomes quite easy. Let us see how we can solve these questions
1. Caustic soda is NaOH, mol wt 40, so 40 g in 1 L will be 1 M solution. You have 4 g in 200 mL or 20 g in 1 L. That is half of 40 g, so molarity will be 0.5 M

mol wt of Na2CO3 is 106. So 106 g/L will be 1 M. You have 5.3 g in 100 mL or 53 g/L, which is again half of its mol wt., so molarity will be 0.5 M once again.

2. Here density is given, which is 1.09 g/mL. So 1 L will weigh 1090 g. Of this sulfuric acid is 13% by mass, or 130 g, rest being water. So weight of water will be 1090 - 130 g = 960 g. So you have 130 g of acid in 960 g of water, which is same as 960 mL of water because water density is 1 g/mL. Now calculate how much acid will be in 1000 ml, which will be 1000*130/960 = 135.4 g. Mol wt of H2SO4 is 98. So divide the weight with its mol wt, or 135.4/98 = 1.38 M.
Now H2SO4 has 2 moles of H+, so the normality will be twice that of molarity, therefore 1.38*2 = 2.76 N

3. Here you express everything on a per 100 g basis. The first oxide contains 27.6 g O or 27.6/16 = 1.725 moles of O and metal will be 100 - 27.6 = 72.4 g. Now the formula given is M3O4, so calculate 4 moles of O will react with how many g of metal, which will be 72.4*4/1.725 = 167.9 g of metal, which is equivalent to 3 moles of metal, so its atomic wt will be 167.9/3 = 55.97 or 56.
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