A chemist adds 295.0 mL of a 0.221M barium chlorate (Ba(ClO3)2) solution to a reaction flask. Calculate the millimoles of barium chlorate the chemist has added to the flask. Round your answer to 3 significant digits.

Respuesta :

Answer:

65.2

Explanation:

For this task, we will use formula C = n/V,

where C is molar concentration (amount of substance per unit volume). The most commonly used unit for molarity is the number of moles per litre with the unit symbol mol/L, or differently M

n is the amount of substance.  The most commonly used unit for the amount of substance is mole with the unit symbol mol

V is the volume of substance. The most commonly used unit for the volume is dm3 which equals to 1L - liter.

In the task we were given the following information:

V = 295.0 mL = 0.2950 L

C = 0.221 M

n = ?

C= n/V ⇒ n = C x V

n = 0.2950 L x 0.221 mol/L

n = 0.065195 mole

We will now transform moles to millimoles because we need to calculate the millimoles of barium chlorate

1 mole = 1000 millimoles ⇒ 0.065195 moles = 65.195

Now, we only need to round this answer to 3 significant digits, so the answer would be 65.2

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