Question & Answer: How many grams of Mg(NO_3)_2 (M_m = 147 g/mol) must be added to 105 g of water to…..

6. 0/8 points| Previous Answers How many grams of Mg (NO3)2 (Mm-147 g/mol) must be added to 105 g of water to make a solution with a colligative molality of 0.195 m? 1.00327 g Mg(NO3)2 Submit Answer Save Progress

How many grams of Mg(NO_3)_2 (M_m = 147 g/mol) must be added to 105 g of water to make a solution with a colligative molality of 0.195 m? g Mg(NO_3)_2

Expert Answer

Answer

let M= mass of Mg(NO3)2 in gm , molar mass of Mg(NO3)2= 147 g/mol, moles of Mg(NO3)2= mass/molar mass= M/147

mass of water= 105 gm, 1000 gm= 1Kg, 105gm= 105/1000Kg= 0.105 Kg

molality= moles of solute/ kg of water

0.195= M/(147*0.105)

M= 0.195*147*0.105 gm =3 gm

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