What mass (in g) of NH_3 must be dissolved in 475 g of methanol to make a 0.250

Question

What mass (in g) of NH_3 must be dissolved in 475 g of methanol to make a 0.250 m solution? 4.94 g 1.90 g 1.19 g 2.02 g 8.42 g Commercial grade HCI solution are typically 39.0% (by mass) HCI in water. Determine the molality of the HCI, if the solution has a density of 1.20 g/mL. 10.7 m 17.5 m 39.0 m 9.44 m 6.39 m Given the term for the amount of solute in moles per liter of solution. mole percent molality mole fraction mass percent molarity Which of the following compounds will be most soluble in ethanol (CH_3CH_2OH)? acetone (CH_3COCH_3) ethylene glycol (HOCH_2CH_2OH) hexane (CH_3CH_2CH_2CH_2CH_2CH_3) trimethylamine (N(CH_3)_3) None of these compounds should be soluble in ethanol. The second-order decomposition of HI has a rate constant of 1.80 - 10^-3 M^-1 s^-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M? 0.258 M 2.39 M 4.55 M 2.20 M 3.87 M Calculate the molarity of a solution that contains 0.250 moles of CsF in 0.500 L of water. 0.750 M 0.500 M 8.00 M 0.125 M 2.00 M A reaction is found to have an activation energy of 38.0 kJ/mol. If the rate constant for this reaction is 1.60 times 10^2 M^-1 s^-1 at 249 K, what is the rate constant at 436 K? 4.20 times 10^5 M^-1 s^1 2.38 times 10^5 M^-1 s^1 7.94 times 10^4 M^1 s^1 3.80 times 10^4 M^-1 s^1 1.26 times 10^3 M^-1 s^-1 Calculate the boiling point of a solution of 500.0 g of ethylene glycol (C_2H_6O_2) dissolved in 500.0 g of water. K_f = 1.86 degree C/m and k_b = 0.512 degree C/m. Use 100 degree C as the boiling point of water. 8.3 degree C 130 degree C 108 degree C 92 degree C 70 degree C What data should be plotted to show that experimental concentration data fits a first-order reaction? In(k) vs. E_a In[reactant] vs. time I/[reactant] vs. time In(k) vs. I/T [reactant] vs. time

Explanation / Answer

(9)

0.250 m solution = 0.25 moles in 1000gm solvent.

number of moles in 475gm solvent (methanol ) = 0.25 mol * 475 g/1000g = 0.1188 moles

1 mol NH3 = molar mass of NH3

So, 1 mol NH3 = 17 g NH3

or, 0.1188 mol NH3 = 17 g * 0.1188 mol/1 mol = 2.02 g

Answer : D

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