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eferences Mailings Review View Normal No spacing Heading 1 Head class independently of the laboratory notebook: Answer to the following questions but rewrite the questions before answering them. 1. Calculate the volume of a 50.0% by mass NaOH solution needed to make 2000. mL of a 0.550M solution. The density of 50.0% by mass NaOH is 1.50 g/mL Show the complete calculation to the correct number of significant figures. (3 points) Hints: 50.00% by mass NaQH means 100. gof solution contains 50.0 g of NaOH. 0.550M NaOH means 1 liter of solution contains 0.550 moles of NaOH, A density of 1 mL means 1.50 g ofNaob per 1 mL ofsolution, Page 4 of 7 CHE 322-L51 Analytical Chemistry Laboratory Spring 2017 (Morse Use these relationships as the conversion factors: 50 n o NaOH 100 o solution

Explanation / Answer

1.

Moleculer weight of NaOH is (23+1+16)=40 . Now, 0.550M NaOH means 1 liter of solution contains 0.550 moles of NaOH i.e (0.550 x 40)=22 g of NaOH . Therefore, 2000 ml (= 2 liter) of 0.550 moles of NaOH solution contains 22 x 2=44 g of NaOH .

Now, 50 % by mass NaOH means, 50 g of NaOH present in 100 g of solution. Then 44 g of NaOH present in [(100 x 44) / 50] = 88 g of NaOH.

Density of the 50% by mass NaOH solution is 1.5 g/ml, i.e 1.5 g of NaOH present in 1 ml of solution. Therefore, 88 g of NaOH present in (88/1.5)=58.67 ml.

So, to make 2000 ml of 0.55 M Naoh solution, 58.67 ml of given 50% by mass NaOh solution is required.

2.

We know, V1S1=V2S2

Here, V1= initial volume

V2=final volume =2000 ml

S1 = initial strength of solution = 10 M

S2= final strength of solution = 0.375 M

Then, V1= [(2000 x 0.375) / 10] = 75 ml

Therefore, to make 2000 ml of 0.375 M HCl solution 75 ml of 10.0 M HCl is required.

3.

KHP is Potassium Hydrogen Phthalate with molecular formula KC8H5O4, and molecular weight 204.22 g/mol. It has only one -CO2H group .Therefore, 1 mole KHP reacts with 1 mole of OH- ion.

Now, 15.75 ml of 0.218 M NaOH is equivalent to [(15.75 x 0.218) / 1000} = 0.00343 mol NaOH which will reacts with 0.00343 mols of KHP.

So, 0.00343 mole of KHP is equivalent to (204.22 x 0.00343) = 0.70 g of KHP.

4.

10.50 ml of 0.305 M HCl is equivalent to [(10.50 x 0.305) / 1000}= 0.0032 mole of HCl

Na2CO3 + 2 HCl = H2CO3 + 2 NaOH

So, 2 mols of HCl react with 1 mole of Na2CO3 . Therefore 0.0032 mole of HCl will react with (0.0032/2)=0.0016 mole of Na2CO3.

Molecular weight of Na2CO3 = [(23x2)+12+(16x3)]=106 g/mol

So, 0.0016 mole of Na2CO3  equivalent to (0.0016 x 106)=0.17 g of Na2CO3

Therefore,0.17 g of Na2CO3 would exactly react with 10.50 ml of 0.305 M HCl.

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