What is the final concentration if 75.0 ml of a 3.50 m glucose solution is diluted to a volume of 400.0 ml?

 the final concentration is 0.656M

Explanation:

The concentration may be expressed in several ways. In this case, since the data talks about volume of the solution, the correct unit of the value 3.50 is M (capital M, and not m), so it is 3.50 M, which means 3.50 Molar

Hence, we are dealing with molarity.

The equation for molarity, M, is:

Wnen you solve for the number of moles of solute, you get:

When it is said that the glucose solution is diluted to a volume of 400.0 ml, means than only solvent was added. That implies, that the number of moles of solute of the final (diluted) solution is the same number of moles of the 3.50 M glucose solution.

That permits you to wite theses equations:

         n₁ = 3.50 M × 75.0 ml × 1 liter/1000 ml

        n₂ = C × 400.0ml × 1 liter / 1000 ml

    ⇒ 3.50 M × 75.0 ml × 1 liter/1000 ml = C × 400.0ml × 1 liter / 1000 ml

        C = 3.50 M × 75.0 / 400.0 = 0.656M

The answer is expressed with three significant digits, because 75.0 ml and 3.50 M are expressed with three significant digits.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol