37th question plzz

37th question plzz AS Calculate acid containing Calculate acid in moles in and the mass per cent Of nitric acid in it tring 40 the molality Of the following solutions (060 g of NaOI'l dissolved in l.Skgofwatcr (ii) 10 g Of 100 g of •aics 41 _ Calculate the mass of urea dissolved in g of water if the molality Of sohnion is Problems based on dilution 42. Conc. HN03 is 14 M. SOO cm3 of this is diluted to S dm'. What is the molarity Ofttrdiluted 43. What is the normality Of a solution prepared by diluting 250 cm3 of0.4 M sulphuric acid wi or water? 44 What volume of concentrated sulphuric acidof molarity 18 M is required to prepare AO solution? 45 Calculate the volume of 10 M HCI required for preparing 20 litres of 0.1 M solution 76 when made up to 250 cm' gives exactb

Dear Student,
As given in the question;
a.) 100 mL (cm³) of 0.1 M oxalic acid (O.A) solution
(considering the oxalic acid to be anhydrous)
I. Molecular weight of O.A = 90.02 g/mol
When oxalic acid is hydrated
II. Molecular weight of O.A = 126 g/mol

0.1 moles = 1000 mL (from molarity given)
1 mL = 0.1/1000 moles
100 mL will have = (0.1/1000) x 100 = 0.01 moles
Case I. Amount of O.A = 0.01 x 90 = 0.9 g
Case II. Amount of O.A = 0.01 x 126 = 1.26 g

b.) Molecular weight of anhydrous sodium carbonate = 106 g/mol
Volume of solution = 500 cm³ or mL
molarity = 0.12 M
Hence 0.12 moles = 1000 mL
So 500 mL = (0.12/1000) x 500 = 0.06 moles
Amount present = 0.06 x 106 = 6.36 grams

Please let me know, if you have any doubts.
Regards

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