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? 494 g 1.90 g 1.19 g 2.02 g 8.42 g Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molality of the HCl, 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 Give 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 he soluble in ethanol. The second-order decomposition of HI has a rate constant of 1.80 times 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 3.57 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? ln(k) vs. E_a ln[reactant] vs. time 1/[reactant] vs. time ln(k) vs. 1/T [reactant] vs. time

Explanation / Answer

Ans 9 D 2.02 grams

Molality is calculated using formula

m = no. Of moles of solute / mass of solvent in kg

Here solute is NH3 and solvent is methanol 475g or 0.475kg and molality is 0.250m

Putting the values in formula, no. Of moles of ammonia will be n = 0.250 × 0.475 = 0.119 moles

Now 1 mole of NH3 = 17g

So 0.119 moles = 0.119 × 17 = 2.02 g

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