Answer:
333g/L
Explanation:
The moles of CaCl2 in 6mEq are:
6mEq * (1mmol / 2mEq) = 3mmol CaCl2.
2Eq / mol because the charge of Ca is 2+
The mass is:
3mmol CaCl2 * (111mg/mmol) = 333mg = 0.333g
Molar mass: 2*35.5g/mol + 40g/mol = 111g/mol = 111mg/mmol
The liters are:
1mL * (1L / 1000mL) = 1x10⁻³L
The concentration in grams per liter is:
0.333g / 1x10⁻³L = 333g/L
