Respuesta :

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

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