What is the molality of a solution that contains 9.16 g of glucose, C_6H_12O_6,

Question

What is the molality of a solution that contains 9.16 g of glucose, C_6H_12O_6, in 228.5 g of water? a. 0.223 m b. 0.0509 m c. 0.00399 m d. 0.0116 m e. 0.0401 m What mass of CaCl_2 in grams is needed to prepare 200 mL of a 0.200 M aqueous CaCl_2 solution? A) 22.1 B) 111 g C) 0.400 D) 4.44 E) none of these In the reaction 2H_2O_2(aq) rightarrow 2H_2O(l) + O_2(g), the initial concentration of H_2O_2 is 0.565 M and, 17.0 seconds later, the concentration of H_2O_2 is 0.361 M. What is the average rate of reaction over this time interval? A. 0.012779 M/s B. 0.12608 M/s C. 0.0212 M/s D. 0.00619 M/s E. 0.00615 M/s What is the K_c equilibrium-concentration of H_2O_2 is 0.565 M and, 17.0 seconds later, the concentration of H_2O_2 is 0.361 M. What is the average rate of reaction over this time interval? A. 0.012779 M/s B. 0.012608 M/s C. 0.0212 M/s D. 0.00619 M/s E. 0.00615 M/s What is the K_c equilibrium-constant expression for the following equilibrium? S_8(s) + 24F_2(g) 8SF_6(g) A) [SF_6]/[S_8][F_2] B) [SF_6]^8/[S_8][F_2]^24 C) [F_2]^24/[SF_6]^8 D) [S_8][F_2]^24 E) [SF_6]^8/[F_2]^24 CS_2 + 3Cl_2 CCl_4(g) + S_2Cl_2(g)

Explanation / Answer

Q12

Molality = mol of solute / kg solvent

mol = mass/M W= 9.16/180 = 0.050888 mol of glucose

kg of solvent = mass/1000 = 228.5/1000 = 0.2285 kg

so

molality = 0.050888/0.2285 = 0.22270 molal

best answer is 0.223 molal, i.e. A)

Q13.

mass of CaCl2 required for

M = 0.2 mol of CaCl2 per liter

V = 200 mL = 0.2 L

mol = M*V = 0.2*0.2 = 0.04 mol of CaCl2

mass = mol*MW = 0.04 *110.98 = 4.4392 g of CaCl2 required

choose D; 4.44 grams

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