At STP, 32 grams of O2 would occupy the same volume as:

At STP 32 g of O₂ would occupy by the same volume as  4 g of He

Complete question

At STP 32 g of O₂ would occupy by the same volume as:

Standard Conditions

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So the gas will have the same volume if the number of moles is the same

mol of 32 grams of O₂ :

So mol of 4 g He = mol of 32 g O₂

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%