The answer is 159.69 g/mol
Answer:
Explanation:
<u>1) Data:</u>
a) M = ?
b) mass of solue = 17 g
c) solute: NH₃
d) V = 0.5o liter
<u>2) Formulae:</u>
a) number of moles, n = mass in grams / molar mass
b) M = n / V (in liters)
<u>3) Solution</u>
a) Molar mass of NH₃ = 17.03 g/mol
b) n = mass in grams / molar mass = 17 g / 17.03 g/mol = 0.998 mol NH₃
c) M = n / V (in liters) = 0.998 mol / 0.50 liter = 1.996 M
d) Round to the appropiate number of significant figures, 2: 2.0 M.
Answer: 2.0 M
You should never taste anything in a lab.
Litmus paper will only tell you whether it is an acid or a base, not how strong it is a hydrogen test only works with acids.
You can get a serious burn from touching strong acids and bases and it doesn't give an objective measure of the strength.
The only one that is left is conductivity. This is safe and rather reliable, because the "strength" of an acid or a base depends on how much it dissociates. Hence more dissociation means greater conductivity.