Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
Answer:
6. d, 7. a
Explanation:
6. Molarity is a number of moles solute in 1 L solution.
7. 1 L solution - 2.5 mol K2CO3
20 L - x mol K2CO3
x =20*2.5/1 = 50 mol K2CO3
Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol
99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.
Actually KCO3 does not exist, in reality it should be K2CO3.
