Suppose you dissolve 2.84 g of succinic acid, C,H(COH), in 500. mL of water Assu

Question

Suppose you dissolve 2.84 g of succinic acid, C,H(COH), in 500. mL of water Assuming that the density of water is 1.00 g/em, calculate the molality of acid in the solution. bSuppose you dissolve 2.84 g of succimic acid, C H(co,HD)2, in 500 ml of water Assuming that the density of water is 100 g/cm calculate the mole fraction of the acid in the solution GSuppose you dissolve 2.84 g of succinic acid, C,Hco,H2 in 500. mL of water Assuming thar the density of water is 1.00gcm calculate the weight percent of the acid in the solution

Explanation / Answer

Ans :number of moles of succinic acid : given mass /118.09

= 2.84 / 118.09

= 0.024 mol

a) Molality = no. of moles of acid / mass of solvent in kg

Since density of water is 1.00 g/cm3, the mass of 500 mL will 500 g

So putting all the values in the formula , we get :

m = 0.024 / 0.500

= 0.048 m

b) The number of moles of water = 500 / molar mass

= 500 / 18.01528

= 27.75

Mole fraction of acid = mol of acid / total moles

= 0.024 / (0.024 + 27.75)

= 8.64 x 10-4

c) weight % = weight of acid / total weight

= [2.84 / ( 2.84 + 500)] x 100

= 0.57 %

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