3. what is the concentration of glucose (c6h1206, mm - 180.16 g/mol) solution that was prepared by
dissolving 50.500 g of glucose in 500.0 ml of solution? assume the solution has a density of 1.00
g/ml.

0.56 M

We are given;

Mass of glucose solute is 50.50 g

Volume of solution is 500 mL

We are required to calculate the concentration of glucose.

We need to know that, molarity is the concentration of a solution in moles per liter.

Molarity = Number of moles ÷ Molar mass

Molar mass of glucose = 180.16 g/mol

Mass of glucose = 50.50 g

Therefore, Moles of Glucose = 50.50 g ÷ 180.16 g/mol

                                                = 0.280 Moles

We can use the formula to calculate the molarity of glucose solution

Moles of glucose = 0.280 moles

Volume of solution = 500 mL or 0.5 L

Molarity = 0.280 moles ÷ 0.5 L

             = 0.56 moles/L

             = 0.56 M

Therefore, the concentration of glucose is 0.56 M

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.