Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
Answer:
0.19 g
Explanation:
Step 1: Given data
Volume of hydrogen at standard temperature and pressure (STP): 2.1 L
Step 2: Calculate the moles corresponding to 2.1 L of hydrogen at STP
At STP (273.15 K and 1 atm), 1 mole of hydrogen has a volume of 22.4 L if we treat it as an ideal gas.
2.1 L × 1 mol/22.4 L = 0.094 mol
Step 3: Calculate the mass corresponding to 0.094 moles of hydrogen
The molar mass of hydrogen is 2.02 g/mol.
0.094 mol × 2.02 g/mol = 0.19 g
The statement that best describes the composition of potassium chlorate, KCIO3 is "<span> The proportion by mass of elements combined in potassium chlorate is fixed."</span>
