What is the concentration of a 0.750 l solution that contains 52.0 g of glucose (c6h12o6)?

1 answer:

7 0

Concentration of Solutions is oftenly expressed in Molarity. Molarity is the number of moles of solute dissolved per volume of solution.

Molarity = Moles / Volume
As,
Moles = Mass / M.mass
So,
Molarity = Mass / M.mass × Volume ---- (1)

Data Given;
Volume = 0.750 L

Mass = 52 g

M.mass = 180 g/mol

Putting Values in eq.1,

Molarity = 52 g ÷ (180 g.mol⁻¹ × 0.750 L)

Molarity = 0.385 mol.L⁻¹

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Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Giv

faltersainse [42]

Answer:

Solution of isopropanol is 10.25 molal

Explanation:

615 g of isopropanol (C3H7OH) per liter

We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.

Density of water 1g/mL → Density = Mass of water / 1000 mL of water

Notice we converted the L to mL

Mass of water = 1000 g (which is the same to say 1kg)

Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles

Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m

4 0

2 years ago

The combustion of propane gas is used to fuel barbeque grills. If 4.65 moles of propane, C3H8, are burned in a grilling session,

Varvara68 [4.7K]

Answer:

<h3>moles of carbon dioxide=13.95mol</h3>

Explanation:

First wrie down the balance chemical reaction:

$C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l)$

Combustion reaction: The reacion in which hydrocarbon is burnt in the presence of oxygen gas and it releases heat and this reaction exothermic because heat of cumbustion is negative.

eg. burning of methane

By using unitry method,

From the above balanced reaction it is clearly that,

from 1 mole of propane 3 moles of carbon dioxide is formed

there fore,

from 4.65 mole of propane $3 \times 4.65$ moles of carbon dioxide will form

moles of carbon dioxide=13.95mol

8 0

2 years ago

A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which

lilavasa [31]

Answer:

$m_{Fe}=12.6gFe$

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

$m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe$