Question

Calculate the amount of reagent and water required to make the following solutions. Do not assume you have a stock solution unless instructed to do so.

a) 1.00 L of 20.00mM Tris-HCl, 50.0mM NaCl, 20.00mM MgSO₄ x 7H₂O, 5.00mM CaCl₂

b) 275 mL of 25% Na₂HCO₃, 18.0% EtOH

c) 750.0mL of 0.2M sodium citrate, 10.00mM SDS (from 0.50M stock), 2.5% glycerol

d) Calculate molarity of 15% NaCl and 27% hydrochloric acid

Answer 1

In first question

Molecular weight of Tris HCl= 121.14 g/mol

According to molarity formula

M=( wt in grams/ Mol wt)/ Volu (Liters)

20 * 10^-3= wt in gra/ 121.14 *1

20 *121.14 * 1 * 10^-3 = 2422.8 * 10^-3= 2.42 gm is required.

For NaCl molecular wt is 58.44 g/mol

50*10^-3 = wt in gm/ 58.44 *1

50 * 58.44 * 1 * 10^-3= 2922 *10^-3= 2.92 gm

Magnesium sulphate seven hydrate molecular wt.

246.47 gm/mol

20 * 10^-3= wt in gram/ 246.47 *1

20 * 246.47 *1 10^-3= 4929.4* 10^-3= 4.92 gm

5mM Calcium chloride Mol wt= 110.98 g/mol

5 * 10^-3= wt in grams/ 110.98 * 1

5* 110.98 *1 10^-3= 554.9 *10^-3= 0.55gm

If individual solution is required then add given amount and make final volume 1lit. if want to make mixture of solution then add all the reagents and make final volume 1 lit.

In second problem, b, for making % solution given amount should be mix.

For 25 % sodium bi carbonate solution

100 mls olution 25 g is require

For 1 ml 25/100

For 275 ml = (275 * 25)/ 100= 68.75 gm is required.

Similarly for ethanol

18 ml is require 100 ml

For 275 ml = (275* 18)/100= 49.5 ml is required.

Individual solution can be made by adding first reagent and make up 275 ml of water.

On the other hand for combined solution

Add 68.75 gm NaHCO₃ and 49.5 ml of ethanol.

Here 275-49.5 ml= ~225.5 ml of water required but final volume should be 275 that will be made in measuring cylinder.

In C question

Mole wt of sodium citrate 214.10 g/mol

0.2 M= wt in gms/ 214.10 * 0.75

0.2 * 214.1* 0.75=32.115 gm is required

10 mM SDS from 0.5 M or 500 mM stock.

10 * 10^-3 *750= 500 10^-3 M * X volume is required

15 ml is required= X

2.5 % glycerol

2.5 gm is required for 100 ml

For 1ml =2.5/100

750 ml= (750 * 2.5)/100

X = 18.75 ml is required

For making combined solution of these three reagent will mix

32.115 gm of citrate

15 ml of SDS Stock and 18.75 ml of glycerol ( 15 +18.75= 33.75)

Approx water required is 750 -33.75= 716.25

But here again due to some wt of powder some less volume is required hence final volume will be 750 ml

In part

15 gm NaCl is dissolved in 100 ml

M = (15 * 1000)/58.55 * 100 ml

M= 0.256 *10

M = 2.56

When volume is in ml then numerator will be multiplied by 1000, that is a conversion factor from liter to ml.

Here calculation is made by W/v ration

Here 27 gm is added to make 100 ml

Molecular wt of HCl= 36..46 g/mol

M = (27 * 1000)/ 36.46 *100

M= 7.4 M