The solubility of a solute and solvent at a given temperature is 30.65 g/100 mL. How much of the solute could be dissolved in 2 L of the solvent?

Please remember to consider significant figure or precise value rules.

613.0 g

Explanation:

2 L = 2000 mL

(30.65 g/100 mL)*2000 mL = 613.0 g

Given:

Solution:

Firstly,

We converted L to mL

2 L = 2 × 1000 mL = 2000 mL [ 1 L = 1000 mL ]

Now,

Solute = 30.65/100 × 2000

= 30.65 × 2000 / 100

= 61,300/100

= 613

its mg hope this

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm