M(H2SO4)=n*M=4.75*98=465.5g
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OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml
</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100
</span>therefore:
V1 = 20 ml
Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.
Answer:
Base Mg(OH)2 does neutralise the acid and is 12g in excess.
Explanation:
2HCL +Mg(OH)2 -> MgCl2 + 2H20
2 * 36.458 g of HCL react with 58.319 g of Mg(OH)2 to neutralise it.
72.916 HCl reacts with 58.319 g of the base.
So 20 g HCl reacts with (58.319/72.916) * 20 = 16g.
There are 28 g of Mg(OH)2 so the base does neutralise all the acid.
The Mg(OH)2 is 28 - 16 = 12 g in excess.
